# Juct check it for me

1. Apr 10, 2004

### moham_87

Hiiiiiiiiiii everyone
I've these two waves
y1=3sin(kx-wt)
y2=4cos(kx+wt)
I need to find the resultant wave (y1+y2)

y= 7 cos(wt+45) sin(kx+45)
is that right?? plz if not give me a hint

==================My efforts=============================
then using trigonometric rules i added the two "sin" and "cos" functions
but i still need to know if my answer is true

bYYyYyyyYYyYYyYE

2. Apr 10, 2004

### Integral

Staff Emeritus
The addition formula that I have for the sum of waves with different amplitudes is:

$$A cos(x) + B sin(x) = \sqrt {A^2 + B^2} cos(x \pm \delta)$$
$$tan( \delta) = \frac {sin \delta} {cos \delta} = \pm \frac B A$$

3. Apr 10, 2004

### moham_87

What is that segma?

4. Apr 10, 2004

### moham_87

What about if "x" is not equal in both equations??

5. Apr 10, 2004

### jdavel

Integral,

The arguments for the sin and cos in your identity are both x. I may be missing a simplification you're seeing, but he's got one wave going left and one going right, so his arguments aren't the same.

I tried using the identities for the sum and difference of two angles, but nothing cancelled, so it just got messy. But like I said, I may be missing something.

6. Apr 10, 2004

### Integral

Staff Emeritus
yeah it gets a little messy but:

$$3 \sin {(kx-\omega t)}= 3( \sin{kx} \cos{\omega t }- \sin{\omega t} \cos {kx} )$$
and
$$4\cos{(kx+\omega t)}= 4(\cos{kx} \cos {\omega t}+ \sin{kx} \sin{\omega t})$$

$$\cos {kx} (4 \cos {\omega t} - 3 \sin {\omega t})+ \sin{kx}(3 \cos {\omega t}+ 4 \sin{\omega t})$$

now apply the formula in my first post to the terms in parentheses.

the $$\delta$$ (thats a low case delta) is defined in the second line of my first post.

Last edited: Apr 10, 2004
7. Apr 10, 2004

### jdavel

Integral, nicely done!

Ok moham87, it doesn't look like your answer's right. But you've got your hint, so have at it! And watch those signs, or they'll kill you.